One kg of copper is drawn into a wire of 1 m | vedclass

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(C) The mass of the copper wire remains constant. Since $Mass = 	ext{Volume} 	imes 	ext{Density} = (A 	imes \ell) 	imes \sigma$, where $A = \pi r

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$16: 1$

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(C) The mass of the copper wire remains constant. Since $Mass = 	ext{Volume} 	imes 	ext{Density} = (A 	imes \ell) 	imes \sigma$, where $A = \pi r^2$ is the cross-sectional area, $\ell$ is the length, and $\sigma$ is the density.For two wires of the same mass: $\pi r_1^2 \ell_1 \sigma = \pi r_2^2 \ell_2 \sigma \implies \frac{\ell_1}{\ell_2} = \left(\frac{r_2}{r_1}ight)^2$.Given diameters $d_1 = 1 \ mm$ and $d_2 = 2 \ mm$, the radii ratio is $\frac{r_2}{r_1} = \frac{2}{1} = 2$. Thus, $\frac{\ell_1}{\ell_2} = (2)^2 = 4$.The resistance $R$ is given by $R = ho \frac{\ell}{A} = ho \frac{\ell}{\pi r^2}$.Therefore, the ratio of resistances is $\frac{R_1}{R_2} = \frac{\ell_1}{\ell_2} 	imes \left(\frac{r_2}{r_1}ight)^2 = 4 	imes (2)^2 = 4 	imes 4 = 16$.Thus, the ratio is $16: 1$.

One $kg$ of copper is drawn into a wire of $1 \ mm$ diameter and a wire of $2 \ mm$ diameter. The resistance of the two wires will be in the ratio:

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